Difficulty solving $\iint \mathrm{ln}(x+{y}^{2})\cdot {y}^{2}\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy$ with change of variables

Sovardipk
2022-07-03
Answered

Difficulty solving $\iint \mathrm{ln}(x+{y}^{2})\cdot {y}^{2}\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy$ with change of variables

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Pranav Greer

Answered 2022-07-04
Author has **13** answers

Let me try to guess. You are interested in a function f:

$\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}x\mathrm{\partial}y}f(x,y)={y}^{2}\mathrm{ln}(x+{y}^{2}).$

You attempt to do a substitution $u=x+{y}^{2}$ and find function $g(u(x,y),v(y))=f(x,y)$

$\frac{\mathrm{\partial}}{\mathrm{\partial}y}(\frac{\mathrm{\partial}}{\mathrm{\partial}x}g(u,v))=\frac{\mathrm{\partial}}{\mathrm{\partial}y}\left(\frac{\mathrm{\partial}g(u,v)}{\mathrm{\partial}u}\frac{\mathrm{\partial}u}{\mathrm{\partial}x}\right)=\frac{\mathrm{\partial}}{\mathrm{\partial}y}\left(\frac{\mathrm{\partial}g(u,v)}{\mathrm{\partial}u}\right)=\frac{{\mathrm{\partial}}^{2}g}{\mathrm{\partial}{u}^{2}}\frac{\mathrm{\partial}u}{\mathrm{\partial}y}+\frac{{\mathrm{\partial}}^{2}g}{\mathrm{\partial}u\phantom{\rule{thinmathspace}{0ex}}\mathrm{\partial}v}\frac{\mathrm{\partial}v}{\mathrm{\partial}y}=\phantom{\rule{0ex}{0ex}}(2y\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{u}^{2}}+2y\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}u\phantom{\rule{thinmathspace}{0ex}}\mathrm{\partial}v})g=2\sqrt{v}(\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{u}^{2}}+\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}u\phantom{\rule{thinmathspace}{0ex}}\mathrm{\partial}v})g.$

So we see that operator $\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}x\mathrm{\partial}y}$ doesn't transform into $\frac{\mathrm{\partial}(u,v)}{\mathrm{\partial}(x,y)}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}u\mathrm{\partial}v}$ the same way as dxdy transforms to $\frac{\mathrm{\partial}(x,y)}{\mathrm{\partial}(u,v)}dudv$. Whereas in 1D $\frac{\mathrm{\partial}}{\mathrm{\partial}q}$ is indeed $\frac{\mathrm{\partial}p}{\mathrm{\partial}q}\frac{\mathrm{\partial}}{\mathrm{\partial}p}$. This is the reason why you can treat derivatives as fractions of differentials in 1D, but cannot do the same in 2D+.

$\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}x\mathrm{\partial}y}f(x,y)={y}^{2}\mathrm{ln}(x+{y}^{2}).$

You attempt to do a substitution $u=x+{y}^{2}$ and find function $g(u(x,y),v(y))=f(x,y)$

$\frac{\mathrm{\partial}}{\mathrm{\partial}y}(\frac{\mathrm{\partial}}{\mathrm{\partial}x}g(u,v))=\frac{\mathrm{\partial}}{\mathrm{\partial}y}\left(\frac{\mathrm{\partial}g(u,v)}{\mathrm{\partial}u}\frac{\mathrm{\partial}u}{\mathrm{\partial}x}\right)=\frac{\mathrm{\partial}}{\mathrm{\partial}y}\left(\frac{\mathrm{\partial}g(u,v)}{\mathrm{\partial}u}\right)=\frac{{\mathrm{\partial}}^{2}g}{\mathrm{\partial}{u}^{2}}\frac{\mathrm{\partial}u}{\mathrm{\partial}y}+\frac{{\mathrm{\partial}}^{2}g}{\mathrm{\partial}u\phantom{\rule{thinmathspace}{0ex}}\mathrm{\partial}v}\frac{\mathrm{\partial}v}{\mathrm{\partial}y}=\phantom{\rule{0ex}{0ex}}(2y\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{u}^{2}}+2y\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}u\phantom{\rule{thinmathspace}{0ex}}\mathrm{\partial}v})g=2\sqrt{v}(\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{u}^{2}}+\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}u\phantom{\rule{thinmathspace}{0ex}}\mathrm{\partial}v})g.$

So we see that operator $\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}x\mathrm{\partial}y}$ doesn't transform into $\frac{\mathrm{\partial}(u,v)}{\mathrm{\partial}(x,y)}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}u\mathrm{\partial}v}$ the same way as dxdy transforms to $\frac{\mathrm{\partial}(x,y)}{\mathrm{\partial}(u,v)}dudv$. Whereas in 1D $\frac{\mathrm{\partial}}{\mathrm{\partial}q}$ is indeed $\frac{\mathrm{\partial}p}{\mathrm{\partial}q}\frac{\mathrm{\partial}}{\mathrm{\partial}p}$. This is the reason why you can treat derivatives as fractions of differentials in 1D, but cannot do the same in 2D+.

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